HDU1711 最基础的kmp算法

Problem Description

Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], ...... , a[K + M - 1] = b[M]. If there are more than one K exist, output the smallest one.

Input

The first line of input is a number T which indicate the number of cases. Each case contains three lines. The first line is two numbers N and M (1 <= M <= 10000, 1 <= N <= 1000000). The second line contains N integers which indicate a[1], a[2], ...... , a[N]. The third line contains M integers which indicate b[1], b[2], ...... , b[M]. All integers are in the range of [-1000000, 1000000].

Output

For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead.

这是一道对kmp算法的最基础运用，在这里主要注意kmp函数以及得到next的函数的写法

代码如下:

 #include <iostream>

 #include <cstdio>

 using namespace std;

 int n,m;

 int a[],b[],next[];

 void getnext (int *s,int *next){

     next[]=next[]=;

     for (int i=;i<m;i++){

         int j=next[i];

         while (j&&s[i]!=s[j])

             j=next[j];

         if(s[i]==s[j]) next[i+]=j+;

         else next[i+]=;

     }

 }

 int kmp (int *a,int *b,int *next){

     getnext (b,next);

     int j=;

     for (int i=;i<n;i++){

         while (j&&a[i]!=b[j])

             j=next[j];

         if (a[i]==b[j])

             j++;

         if (j==m)

             return i-m+;

     }

     return -;

 }

 int main (){

     int t;

     scanf ("%d",&t);

     while (t--){

         scanf ("%d %d",&n,&m);

         for (int i=;i<n;i++)

             scanf ("%d",&a[i]);

         for (int i=;i<m;i++)

             scanf ("%d",&b[i]);

         int ans=kmp (a,b,next);

         printf ("%d\n",ans);

     }

     return ;

 }

巴特西

HDU1711 最基础的kmp算法

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